CODEWITHSUNDEEP
pythonpandas
zork
zork

Reputation: 2135

Python Pandas: differences between two dates in weeks?

When trying to find differences between two dates in weeks:

import pandas as pd

def diff(start, end):
    x = millis(end) - millis(start)
    return x / (1000 * 60 * 60 * 24 * 7 * 1000)

def millis(s):
    return pd.to_datetime(s).to_datetime64()

diff("2013-06-10","2013-06-16")

As a result I get:

Out[15]: numpy.timedelta64(857,'ns')

Which is obviously wrong. Questions:

  1. How to get the difference in weeks, not nanoseconds, rounded up to a whole value?

  2. How to get value out of 'numpy.timedelta64' object?

Upvotes: 8

Views: 12477

Answers (4)

b.ill
b.ill

Reputation: 1

jezrael's answer threw an error for me so here's an alternate solution (in case you also got an error when trying it)

import numpy as np
import pandas as pd
def diff(start, end):
    x = pd.to_datetime(end) - pd.to_datetime(start)
    return (x).apply(lambda x: x/np.timedelta64(1,'W')).astype(int)

Upvotes: 0

nimbous
nimbous

Reputation: 1527

You can use pandas.Timedelta as well:

import pandas as pd

def diff(start, end):
    days = pd.to_datetime(end) - pd.to_datetime(start)
    week = int(pd.Timedelta(days).days / 7)
    remainder = pd.Timedelta(days).days % 7
    return str(week) + ' weeks and ' + str(remainder) + ' days'

print(diff("2019-06-10","2019-07-11"))

Output:
4 weeks and 3 days

Upvotes: 2

jezrael
jezrael

Reputation: 862581

I think you can convert to int by dividing by numpy scalar:

def diff(start, end):
    x = pd.to_datetime(end) - pd.to_datetime(start)
    return int(x / np.timedelta64(1, 'W'))

print (diff("2013-06-10","2013-06-16"))
0
print (diff("2013-06-10","2013-06-26"))
2

See frequency conversion.

Upvotes: 14

John Zwinck
John Zwinck

Reputation: 249133

Here's a simple fix:

def diff(start, end):
    x = millis(end) - millis(start)
    return np.ceil(x.astype(int) / (7*86400*1e9))

The main thing is to remove the units (nanoseconds) before operating on it.

P.S.: Consider not calling your function millis() when it does not return milliseconds.

Upvotes: 1

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